User blog:Rgetar/Veblen-like collapsing function up to Buchholz ordinal
This is modified version of my Veblen-like functions. Modifications Subscripts Now φαβ + 1 + γ(X) = φαβ(φα + 1 + βγ(X)) (Edited version: φαβ + 1(X) = φαβ(φα + 1 + β(X)) ) (Old version: φαγ(X) = φαβ(φβγ(X)) ) And now φα(X) = φα0(X) (Old version: φα(X) = φαα + 1(X) ) So, φα(X) = φα0(X) φα(X) = φ0α(X) φ(X) = φ00(X) Advantage of the new version: I think, it should be easier for ordinals beyond Buchholz ordinal φω(0). Small arguments Now for X < Ωα + 1 φα(X): φα(X) = ωA + X where A is such as φα(0) = Ωα That is φ(0) = ωA = Ω0 = ω, so A = 1 For α > 0 φα(0) = ωA = Ωα, so A = Ωα That is φ(X) = ω1 + X for X < Ω φα(X) = ωΩα + X for X < Ωα + 1, α > 0 Or φα(X) = ΩαωX for X < Ωα + 1 (Old version: φα(X) = ΩαX for X < Ωα + 1) Advantages of the new version. There are some advantages: *Now cardinality of φαβ(X) is always Ωα *Simple expression for Ωα: Ωα = φα(0) *Now we do not need multipliers for Cantor normal form terms, for example, Ωω = φ1(1), but in old version φ1(1) = Ω, φ1(2) = Ω2, so we need multiplier ω to express Ωω Definition Definition of Veblen-like collapsing function (VCF) φαβ(X) for β < ω: Range of input for φαβ(X): X < Ωα + 1 + β + 1 If α or β in φαβ(X) is 0, then we can omit it: φα(X) ≡ φα0(X) φα(X) ≡ φ0α(X) (I use this definition of cofinality: cofinality cof(α) of ordinal α is minimal length of increasing sequence such as α is minimal ordinal larger than all elements of this sequence. That is cof(1) = 1). (And I use this definition of fundamental sequence: fundamental sequence of ordinal α is strictly increasing sequence of length cof(α) such as α is minimal ordinal larger than all elements of this sequence. That is 10 = 0). Infinite ordinals are defined using fundamental sequences. Cofinality of ordinal is cofinality of its last Cantor normal form term. n-th element of fundamental sequence of ordinal γ is Cantor normal form of γ with last term, replaced with n-th element of fundamental sequence of this term. Any Cantor normal form term in this notation is 1 or φαβ(X). φαβ + 1(X) = φαβ(φα + 1 + β(X)) Definition: X0 is X without all Cantor normal form terms of cardinality less than card(X). Six cases Six cases of Cantor normal form term φα(X): #φα(0) #φα(X + 1), X < Ωα + 1 #φα(X + 1), X ≥ Ωα + 1, ω ≤ cof(X0) ≤ Ωα #φα(X + 1), X ≥ Ωα + 1, cof(X0) = Ωα + 1 #φα(X), ω ≤ cof(X) ≤ Ωα #φα(X), cof(X) = Ωα + 1 Cofinality Corresponding cofinalities of term: #Ωα #ω #cof(X0) #ω #cof(X) #ω Fundamental sequence Corresponding n-th elements of fundamental sequence of term: #n #φα(X)n #φα(X0n + φα(X) + 1) #φα(X0+ 1) for n = 0, φα(X0+ 1)[n - 1]) for n > 0 #φα(Xn) #φα(X0) for n = 0, φα(X[φα(X)- 1]) for n > 0 Standard forms φαβ(X) is in standard form, if it does not contain ordinal equal to it as its part. Otherwise it is in non-standard form. Note: a standard form can contain non-standard form as its part, and this part should not be converted into standard form, otherwise we can get different ordinal. That is we convert into standard form only ordinal itself, not its parts (we can convert parts, but then we also should check, if ordinal itself changed). Example: let φ(φ1(X)) and φ1(X) are standard forms. And let φ1(Y) is non-standard form of φ1(X), containing countable part larger than φ(φ1(X)). If we replace φ1(X) with φ1(Y) in φ(φ1(X)), then we get ordinal larger than φ(φ1(X)), despite φ1(Y) = φ1(X). And this larger ordinal φ(φ1(Y)) can be standard form, despite φ1(Y) is non-standard form. (See also my previous blog Three properties of standard forms). Computer format Computer format is format of VCF, used by my program. In computer format five symbols are used: "1", "+", "(", ")", "." (In old version of computer format there were multipliers, but now we do not need them). Empty string is 0. 1 is 1. Cantor normal form terms are separated with "+". (α.β.X) = φαβ(X) (β.X) = (.β.X) = φβ(X) (X) = (..X) = φ(X) Examples '''' = 0 1 = 1 1+1 = 2 1+1+1 = 3 1+1+1+1 = 4 1+1+1+1+1 = 5 () = ω ()+1 = ω + 1 ()+() = ω2 ()+()+1 = ω2 + 1 (1) = ω2 (1+1) = ω3 (1+1+1) = ω4 (1+1+1+1) = ω5 (()) = ωω (()+1) = ωω + 1 (()+()) = ωω2 ((1)) = ωω2 ((())) = ωωω ((((())))) = ωωωωω ((1..)) = (1.) = ε0 ((1..)+1) = ε1 ((1..)+(1..)) = ζ0 ((1..)+(1..)+(1..)) = η0 ((1..1)) = φ(ω, 0) ((1..())) = φ(ωω, 0) ((1..((1..)))) = φ(ε0, 0) ((1..(1..))) = Γ0 ((1..(1..)+(1..))) = φ(1, 0, 0, 0) ((1..(1..1))) = Small Veblen ordinal ((1..(1..(1..)))) = Large Veblen ordinal ((1..(1+1..))) = (1+1.) = Bachmann-Howard ordinal (().) = Buchholz ordinal (1..) = Ω (1+1..) = Ω2 (1+1+1..) = Ω3 (1+1+1+1..) = Ω4 (1+1+1+1+1..) = Ω5 Category:Blog posts